2 Answers
2

Assume that $X$ and $Y$ are independent, that is
$$
P(X\in A,Y\in B)=P(X\in A)P(Y\in B),\quad \text{for all}\; A,B\subseteq\mathbb{R}.\tag{1}
$$
To show that $g(X)$ and $Y$ are independent, we have to show $(1)$ with $X$ replaced by $g(X)$.

Note that in general we can only talk about the probability of $P(X\in A)$ for certain nice subsets $A$. These are called Borel sets, and hence $(1)$ is only required to hold for Borel sets $A$ and $B$. But this is probably out of the scope of your question.

If the joint density function of $X$ and $Y$ is the product of the individual densities then they are considered independent. Therefore, $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ the $X$ and $Y$ are independent. Let $Z=f(x)$. You have to show that:

$$f_{Z,Y}(z,y) = f_Z(z) f(y)$$

Start from the joint CDF of $Z$ and $Y$ and see if you can work it out.

Thanks for your answer, Im not familiar with the notation you used though ($f_{X,Y}(x,y)=f_X(x)f_Y(y)$), what does $f_{X,Y}(x,y)$ mean? Sorry if this sounds stupid but i think my course may use different notation! I am in uk if that is relevent.....Thanks
–
Bernard.MathewsOct 23 '13 at 3:46

The notation is used to indicate that the functions are potentially different. For example, $X$ can be a normal pdf and $Y$ can be a uniform pdf. In this case, $f_X$ refers to the normal and $f_Y$ refers to the uniform and $f_{X,Y}$ refers to the joint density function. Since, I am using the same symbol $f$, I need a way to communicate the idea that the pdfs of $X$, $Y$ and their joint all are potentially different functions.
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responseOct 23 '13 at 3:51

Oh okay, i kind of understand, so do i have to show that $P(X=x, Y=y)=P(X=x)P(Y=y)$?
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Bernard.MathewsOct 23 '13 at 3:53

No, those probabilities would be $0$ as you have continuous random variables. Start from $P(Z \le z, Y \le y) = P(g(X) \le z, Y \le y)$. Then invert $g(X)$ to re-write the last probability, then use independence of $X$ and $Y$ and see if you can progress.
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responseOct 23 '13 at 3:57

Ahh im completely confused!! This notation is not used in my lecture notes or model solutions! :/
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Bernard.MathewsOct 23 '13 at 4:00